This set of Prestressed Concrete Structures Multiple Choice Questions & Answers (MCQs) focuses on “Short Term Deflections”.

1. The short term deflections are also known as:

a) Cracked

b) Un cracked

c) Instantaneous

d) Non instantaneous

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Explanation: Short term deflections of prestressed members are also known as instantaneous deflections governed by distribution of bending moment throughout the span and flexural rigidity of member, these theorems are applied for determining the deflections due to prestressing force, imposed loads and self weight.

2. Which of the following is the equation given Mohr’s first theorem?

a) Area of bending moment deflection/flexural rigidity

b) Moment/flexural rigidity

c) Deflection/flexural rigidity

d) Loads/flexural rigidity

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Explanation: When the beam AB is subjected to a bending moment distribution due to prestressing force or self weight or imposed loads, ACB is the centre line of the deformed structure under the system of given loads, According to Mohr’s first theorem

Slope = area of bending moment deflection/flexural rigidity, θ = A/EI.

3. Which of the following is the equation given by Mohr’s second theorem?

a) Mid span/flexural rigidity

b) Moment of area of bending moment diagram/flexural rigidity

c) End span/flexural rigidity

d) Thickness/flexural rigidity

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Explanation: Mohr’s second theorem states that

Intercept, a = (moment of the area of bending moment deflections/flexural rigidity), a = AX/EI,

a = deflection at the centre for symmetrically loaded, simply supported beam (since the tangent is horizontal for such cases), A = area of bending moment deflection between A and C, x = distance of the centroid of the bending moment deflection between A and C from the left support, EI = flexural rigidity of beam.

4. Which of the following deflections are directly obtained by Mohr’s second area theorem?

a) Simply supported beam

b) Uniformly distributed load

c) Point beams

d) Fixed beams

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Explanation: The deflections of symmetrically loaded and simply supported beam at the mid span point are directly obtained from the second moment area theorem since the tangent is horizontal at this span, In most of cases of prestressed beams tendons are located with eccentricities towards the soffit of the beam to counteract the sagging bending moments due to transverse loads.

5. The problems involving unsymmetrical loading can be solved by:

a) Mohr’s theorem

b) Kennedy’s theorem

c) Row’s theorem

d) Casagrande’s theorem

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Explanation: More complicated problems involving unsymmetrical loading may be solved by combining both the moment area theorems Mohr’s first theorem and second theorem, since the bending moment at every section is the product of prestressing force and eccentricity the tendon profile itself will represent the shape of the bending moment diagram.

6. A straight tendon at a uniform eccentricity below the centroidal axis is given as:

a) –PeL^{2}/4EI

b) –PeL^{2}/8EI

c) –PeL^{2}/14EI

d) –PeL^{2}/16EI

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Explanation: A straight tendon at a uniform eccentricity below the centroidal axis is given as:

If the camber of beam with straight tendons upward deflections are considered as negative and

a = -(PeL) (L/4)/EI = -PeL

^{2}/8EI, P = effective prestressing force, e = eccentricity, L = length of beam.

7. A tendon with a trapezoidal profile considering the bending moment and deflection at the centre of the beam is obtained by:

a) –Pe/6EI(2l_{1}^{2}+6l_{1}l^{2}+3l_{2}^{2})

b) –Pe/6EI(2l_{1}^{2}+6l_{1}l^{2}+3l_{2}^{2})

c) –Pe/6EI(2l_{1}^{2}+6l_{1}l^{2}+3l_{2}^{2})

d) –Pe/6EI(2l_{1}^{2}+6l_{1}l^{2}+3l_{2}^{2})

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Explanation: A draped tendon with a trapezoidal profile considering the bending moment diagram the deflection at the centre of the beam is obtained by taking the moment of the area of the bending moment diagram over one half of the span A = –Pe/6EI(2l

_{1}

^{2}+6l

_{1}l

^{2}+3l

_{2}

^{2}).

8. The deflection of a beam with parabolic tendon is given as:

a) –5PeL^{2}/48EI

b) –10PeL^{2}/48EI

c) –15PeL^{2}/48EI

d) –3PeL^{2}/48EI

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Explanation: The deflection of the beam with parabolic tendons having an eccentricity e at the center and zero at the supports is given by a = –5PeL

^{2}/48EI, a beam with a parabolic tendon having an eccentricity e

_{1}at the centre of span and e

_{2}at the support sections and the resultant deflection at the centre is obtained as the sum of the upward deflection of a beam with a parabolic tendon of eccentricity e

_{1}+e

_{2}at the centre and zero at the supports and the downward deflection of a beam subjected to a uniform sagging bending moment of intensity pe

_{2}throughout the length, the resultant stress becomes a = PL

^{2}/48EI(-5e

_{1}+e

_{2}).

9. The deflection is computed in a way similar to sloping tendon is given as:

a) 2PL^{2}/24EI

b) 4PL^{2}/24EI

c) PL^{3}/24EI (-2e_{1}+e_{2})

d) PL^{2}/24EI (e_{1}+e_{2})

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Explanation: The deflection in sloping tendon is computed in a way similar to:

A = (-PL

^{2}/12EI(e

_{1}+e

_{2})) + (Pe2L

^{2}/8EI)

A = (PL

^{3}/24EI (-2e

_{1}+e

_{2})).

10. The deflection due to self weight and imposed loads are:

a) 5(g+q)L^{4}/384EI

b) 5(g+q)L^{4}/384EI

c) 5(g+q)L^{4}/384EI

d) 5(g+q)L^{4}/384EI

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Explanation: At the time of transfer of prestress, the beam hogs up due to the effect of prestressing, at this stage the self weight of the beam includes downward deflections, which further increases due to the effect of imposed loads on the beam a = 5(g+q)L

^{4}/384EI and deflections due to concentrated live loads can be directly computed by using Mohr’s theorem.

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